Ricevo da Lucia la seguente domanda:
Carissimo professore,
ho delle difficoltà con alcuni limiti (pag.1522, nn. 169, 176, 177, 178, Matematica.blu 2.0)
\[\underset{x\to -\infty }{\mathop{\lim }}\,\frac{3x-2}{\sqrt{{{x}^{2}}-x+1}}\quad \underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+2{{x}^{2}}-1}{\sqrt{{{x}^{6}}+3{{x}^{2}}+2}}\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{9{{x}^{4}}+5{{x}^{2}}+x}}{{{\left( x+2 \right)}^{2}}}\quad \underset{x\to +\infty }{\mathop{\lim }}\,\frac{5{{x}^{2}}-1}{4+{{x}^{2}}}\sqrt{\frac{2-3x}{1-12x}}\quad .\]
Grazie.
Le rispondo così:
Cara Lucia,
premettiamo un semplice fatto algebrico: posto che sia \({{x}^{2n}}\), con \(n\in \mathbb{N}\), una potenza pari di \(x\), distinguamo due possibilità: \[n=2k\to \sqrt{{{x}^{2\left( 2k \right)}}}=\sqrt{{{x}^{4k}}}={{x}^{2k}}\quad \vee \quad n=2k+1\to \sqrt{{{x}^{2\left( 2k+1 \right)}}}=\sqrt{{{x}^{2}}\cdot {{x}^{4k}}}=\left| x \right|{{x}^{2k}}\quad .\] Alla luce di questo, e ricordando che: \[\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left| x \right|}{x}=1\quad \underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left| x \right|}{x}=-1\] si ha: \[\underset{x\to -\infty }{\mathop{\lim }}\,\frac{3x-2}{\sqrt{{{x}^{2}}-x+1}}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{3x\left( 1-\frac{2}{3x} \right)}{\left| x \right|\sqrt{1-\frac{1}{x}+\frac{1}{{{x}^{2}}}}}=3\cdot \underset{x\to -\infty }{\mathop{\lim }}\,\frac{x}{\left| x \right|}\cdot \underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left( 1-\frac{2}{3x} \right)}{\sqrt{1-\frac{1}{x}+\frac{1}{{{x}^{2}}}}}=3\cdot \left( -1 \right)\cdot 1=-3\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+2{{x}^{2}}-1}{\sqrt{{{x}^{6}}+3{{x}^{2}}+2}}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}\left( 1+\frac{2}{x}-\frac{1}{{{x}^{3}}} \right)}{\left| x \right|{{x}^{2}}\sqrt{1+\frac{3}{x^4}+\frac{2}{{{x}^{6}}}}}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x}{\left| x \right|}\cdot \underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left( 1+\frac{2}{x}-\frac{1}{{{x}^{3}}} \right)}{\sqrt{1+\frac{3}{x^4}+\frac{2}{{{x}^{6}}}}}=\left( -1 \right)\cdot 1=-1\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{9{{x}^{4}}+5{{x}^{2}}+x}}{{{\left( x+2 \right)}^{2}}}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{3\sqrt{1+\frac{5}{9{{x}^{2}}}+\frac{1}{9{{x}^{3}}}}}{x\left( 1+\frac{4}{x}+\frac{4}{{{x}^{2}}} \right)}=3\cdot 1=3\] \[\underset{x\to +\infty }{\mathop{\lim }}\,\frac{5{{x}^{2}}-1}{4+{{x}^{2}}}\sqrt{\frac{2-3x}{1-12x}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{5\left( 1-\frac{1}{5{{x}^{2}}} \right)}{\left( 1+\frac{4}{{{x}^{2}}} \right)}\cdot \underset{x\to +\infty }{\mathop{\lim }}\,\sqrt{\frac{3\left( 1-\frac{2}{3x} \right)}{12\left( 1-\frac{1}{12x} \right)}}=5\cdot \frac{1}{2}=\frac{5}{2}\quad .\]
Massimo Bergamini